Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e4 + 5;
int n, a[N], b[N];
vector<pair<int, int> > ans1, ans2;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= n; i++) {
    cin >> b[i];
  }
  for (int i = 1, j = 30; i <= n && j >= 0; j--) {
    for (int k = i; k <= n; k++) {
      if (a[k] & (1 << j)) {
        if (i == k) {
          break;
        }
        a[i] ^= a[k];
        ans1.push_back(make_pair(i, k));
        break;
      }
    }
    if ((a[i] & (1 << j)) == 0) {
      continue;
    }
    for (int k = 1; k <= n; k++) {
      if (k != i && (a[k] & (1 << j))) {
        a[k] ^= a[i];
        ans1.push_back(make_pair(k, i));
      }
    }
    i++;
  }
  for (int i = 1, j = 30; i <= n && j >= 0; j--) {
    for (int k = i; k <= n; k++) {
      if (b[k] & (1 << j)) {
        if (i == k) {
          break;
        }
        b[i] ^= b[k];
        ans2.push_back(make_pair(i, k));
        break;
      }
    }
    if ((b[i] & (1 << j)) == 0) {
      continue;
    }
    for (int k = 1; k <= n; k++) {
      if (k != i && (b[k] & (1 << j))) {
        b[k] ^= b[i];
        ans2.push_back(make_pair(k, i));
      }
    }
    i++;
  }
  for (int i = 1; i <= n; i++) {
    int x = a[i], y = b[i];
    for (int j = 30; j >= 0; j--) {
      if (((x ^ y) >> j) & 1) {
        for (int k = 1; k <= n; k++) {
          if ((a[k] >> j) & 1) {
            ans1.push_back(make_pair(i, k));
            x ^= a[k];
            break;
          }
        }
      }
    }
    if (x != y) {
      return cout << "-1", 0;
    }
  }
  cout << ans1.size() + ans2.size() << "\n";
  for (auto it : ans1) {
    cout << it.first << " " << it.second << "\n";
  }
  reverse(ans2.begin(), ans2.end());
  for (auto it : ans2) {
    cout << it.first << " " << it.second << "\n";
  }
}